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Figure 5-22. Digital to Analog Encoding. 3 Characteristics of a sine wave Amplitude Frequency Phase 3 Mechanisms for Modulating Digital Data into Analog Signal ASK FSK PSK. Figure 5-23. Bit Rate and Baud Rate. Bit rate number of bits transmitted during 1 second Baud rate

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## Figure 5-22

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**Figure 5-22**Digital to Analog Encoding**3 Characteristics of a sine wave**• Amplitude • Frequency • Phase • 3 Mechanisms for Modulating Digital Data into Analog Signal • ASK • FSK • PSK**Bit Rate and Baud Rate**• Bit rate • number of bits transmitted during 1 second • Baud rate • number of signal units per second • determines the BW required to send the signal Bit rate = baud rate * # of bits represented by each signal unit**Bit Rate and Baud Rate**• An analog signal carries four bits in each signal element. If 1000 signal elements are sent per second, find the baud rate and the bit rate. • The bit rate of a signal is 3000. If each signal element carries six bits, what is the baud rate?**ASK**Each shift represents a single bit**Figure 5-25**Bandwidth for ASK BW of a signal is the total range of frequencies occupied by that signal Nbaud = baud rate fc = carrier frequency**Find the minimum BW for an ASK signal transmitting at 2000**bps. The transmission mode is half-duplex • Solution: • In ASK bit rate and baud rate are the same. • An ASK signal requires a min BW equal to its baud rate**Given a BW of 10,000 Hz (1000 – 12,000 Hz), draw the**full-duplex ASK diagram of the system. Find the carriers and the BWs in each direction. Assume there is no gap between the bands in 2 directions • Solution: BW for each direction = 10,000/2 = 5000 Hz The carrier frequencies can be chosen at the middle of the bands fc(forward) = 1000 + 5000/2 = 3,500Hz fc(backward) = 11,000 - 5000/2 = 8,500Hz**Figure 5-27**FSK**Figure 5-28**Bandwidth for FSK**Find the minimum BW for an FSK signal transmitting at 2000**bps. The transmission mode is half-duplex and the carriers must be separated by 3,000 Hz • Solution: BW = baud rate + (fc1 –fc0) The baud rate is the same as the bit rate BW = 2000 + 3000 =5000 Hz**Find the max bit rates for an FSK signal if the BW of the**medium is 12,000 Hz and the difference between the carriers must be at least 2000 Hz. Transmission is in full-duplex mode. • Solution: BW = baud rate + (fc1 –fc0) The BW for each direction is 6000 Hz Baud rate = 6000 –2000 = 4000 Baud rate = bit rate Bit rate = 4000 bps**Figure 5-29**PSK**Figure 5-30**PSK Constellation**Figure 5-31**4-PSK**Figure 5-32**4-PSK Characteristics**Figure 5-33**8-PSK Characteristics**Figure 5-34**PSK Bandwidth The minimum BW for PSK transmission is the same as that required for ASK transmission Max baud rates of ASK and PSK are the same for a given BW, but the bit rates could be 2 or more times greater**(1) Find the BW for a 4-PSK signal transmitting at 2000 bps.**Transmission is in half-duplex mode (2) Given the BW of 5000Hz for an 8-PSK signal, what are the baud and bit rate?**(1) Solution:**Baud rate is half of the bit rate. A PSK signal requires a BW equal to its baud rate (2) Solution: Baud rate = 5000 • Bit rate = 3 (5000) = 15, 000 bps**QAM**Quadrature Amplitude Modulation • Combined ASK and PSK • If there are x variations in phase and y variations in amplitude, it will give us x times y possible variations.**Figure 5-35**4-QAM and 8-QAM Constellations 4 possible variations 8 possible variations**Figure 5-36**Time domain for 8-QAM Signal**16-QAM Constellation – Different configurations**16 out of 36/32 possible variations are utilized – to ensure readability Greater ratio of phase shift to amplitude handles noise best 1st figure – ITU-T recommendation 2nd figure – OSI recommendation**16-QAM Constellation – Different configurations**Several QAM design link specific amplitudes with specific phases. This means that even with the noise problems associated with amplitude shifting, the meaning of the shift can be recovered from phase information**Figure 5-38**Bit Rate and Baud Rate**Figure 5-38-continued**Bit Rate and Baud Rate**EXAMPLES**• A constellation diagram consists of 8 equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? (2) Compute the bit rate for a 1000-baud 16-QAM signal. (3) Compute the baud rate for a 72,000 bps 64-QAM signal.**(4) The data points of a constellation are at (4,0) and**(6,0). Draw the constellation. Show the amplitude and phase for each point. Is the modulation ASK, PSK, or QAM? How many bits per baud can one send with this constellation? (5) Repeat the exercise above if the data points are (4,5) and (8,10)**(1) Solution:**• the constellation indicates 8-PSK with the points 45 degrees apart • 3 bits are transmitted with each signal element • 4800/3 = 1600 baud

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